Question

The sides $a,b,c$ of $△ABC,$ are in A.P.If $\cos \alpha =\frac{a}{b+c},\cos \beta =\frac{b}{c+a},\cos \gamma =\frac{c}{a+b}$ then ${\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\gamma }{2}=$

• A. $1$
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. $\frac{2}{3}$

Answer 1

The correct option is D $\frac{2}{3}$

Given:$a,b,c$ are in A.P
$\Rightarrow 2b=a+c$
$\frac{\cos \alpha }{1}=\frac{a}{b+c}$
By componendo-dividendo rule, we have
$\frac{1-\cos \alpha }{1+\cos \alpha }=\frac{b+c-a}{b+c+a}$
$\Rightarrow {\tan }^2\frac{\alpha }{2}=\frac{b+c-a}{3b}$ where $a+c=2b$
$\frac{\cos \gamma }{1}=\frac{c}{a+b}$
By componendo-dividendo rule, we have
$\frac{1-\cos \gamma }{1+\cos \gamma }=\frac{a+b-c}{a+b+c}$where $a+c=2b$
$\Rightarrow {\tan }^2\frac{\gamma }{2}=\frac{a+b-c}{3b}$
$\therefore {\tan }^2\frac{\alpha }{2}+{\tan }^2\frac{\beta }{2}=\frac{b+c-a+a+b-c}{3b}=\frac{2b}{3b}=\frac{2}{3}$